poj-2299-Ultra-QuickSort

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Description

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence 
9 1 0 5 4 ,
Ultra-QuickSort produces the output 
0 1 4 5 9 .
Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

Input

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.

Output

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.

Sample Input

5
9
1
0
5
4
3
1
2
3
0

Sample Output

6
0

解题思路：

这个题就不能像之前那样直接求逆序对了，一定会超时(o(n^2))。

采用归并算法的原理，复杂度降到o(nlog(n))，具体分析可百度，cnt=mid-i+1。

#include<stdio.h>
# define N 500050
int a[N];
long long int cnt;
void merge(int l,int mid,int r)
{
   int i=l,j=mid+1,k=l;//
   int tmp[N];

   while(i<=mid&&j<=r)
   {
     if(a[i]<=a[j])
	 {
		tmp[k++]=a[i++];
	 }
	 else
	 {
		 tmp[k++]=a[j++];
		 cnt+=mid-i+1;//求逆序对
	 }
   }

   while(i<=mid)
	   tmp[k++]=a[i++];
   while(j<=r)
	   tmp[k++]=a[j++];

   for(i=l;i<=r;i++)//
	   a[i]=tmp[i];
}

void merge_sort(int l,int r)
{
   if(l<r)
   {
      int mid=(l+r)/2;
	  merge_sort(l,mid);
	  merge_sort(mid+1,r);

      merge(l,mid,r);
   }
}
int main()
{

	int n;
	while(scanf("%d",&n)!=EOF&&n)
	{
	  int i;
	  for(i=1;i<=n;i++)
	     scanf("%d",&a[i]);


      cnt=0;   
      merge_sort(1,n);
	  printf("%lld\n",cnt);

	}

return 0;
}